Question

The geometric distribution is a probability distribution of the number X of Bernoulli trials needed to get one success. For example, how many attempts does a basketball player need to get a goal. Given the probability of success in a single trial is p, the probability that the xth trial is the first success is: Pr(x = x|p) = (1 - p*-'p for x=1,2,3,.... Suppose, you observe n basketball players trying to score and record the number of attempts required to get the first goal: X1, X2, ..., Xn for each of them. (a) Write down your likelihood and find the MLE of p. (1pt) (b) Using Bayesian approach, assign a beta prior to p and obtain the posterior distribution for p. (1pt) (c) Compare the posterior mean of p to the MLE. What can you say about sensitivity of posterior inference to prior assumptions? What happens to the difference between the posterior mean and the MLE of p as the sample size n increases? (1pt)

Answer 1

Solutions Given that: a ..un OS i =) PX [x = xlp J = (1-P3*"'p let n, x2, ²3, be scindom Sample from Bernoulli 1P) distribution with pinoy PC x=-] = (1-P)ap 1=1,2,3...-- las The likelihood function Jos p as [[Pla3 = ✓ P{*=-17] (1-P) *;-) ? = [1-p] Exianph The logy likelihood fonchón jos pas loge [plus] = [xx.n] log (17) + log p differentiate withrespect with pl we get dlog (plu] . [Eni-n] to dp cit] gp we get top FOR MLE op 1094[P/x] + 50 [{xi-n 1-P n = {xi-n n(1-P) - PZx; -op anpa pening n =P

n Eni iel n Hence the MLE of p as E ni b, 1 0<p<1 Let the prior for p as. TT (P) pd-1 (1-P) B-1 Beta(d,p) where ao & ß>O theolem Then the posterios distribution of p’using Bayes We get { xi-n + B-1 зо<p<) ī (Plx) d pntdy (1-P) .:|[Plx)= pn td-1 (1-P) E 11-9 + B - Beta (ntd {n; -n+B Hence the posterios distribution Plan Beta Cnto, Eni-n+B] as mean then, Po= E(Plu] C) t let PB be the posterior JP TCP/x] dp sonte (1-P) $ *; -0+ B-1.dp Bet a[n+d, 3 m;-N+B]ó 1 Beta (1 +n+d, Ex; -9+1] - In tot Bet a [n+d, Eli-n+B] Şi-n+B Enitatß Elitatßt Inta Texante

Entd faith+ Thus posterios mean as ntd Exitat Now, we have 71 n <a ६ PB And nntatß - 1+010 ā +(2+3)]n Then it f = ai la りますの 0700 & PB itolo 5+(2+3) WEq it n=0 1X > 10889 . We get from 29 (VG 29 () at PB it - n 기 n-10 n) - 0 mean & MIE g If noo then PB = * There is difference between the posterior p as the sample size 'n' increases no